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NEET-XII-Physics

33: Thermal and Chemical Effects of Electric Current

with Solutions - page 5

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  • Qstn #14
    Find the thermo-emf developed in a copper-silver thermocouple when the junctions are kept at 0°C and 40°C. Use the data in table (33.1).
    Ans : Difference in temperature, θ = 40°C
    Emf, Ecs = acsθ + `` \frac{1}{2}``bcsθ2 ...(1)
    acs = [2.76 - (-43.7) μV
    = 46.46 μV/°C
    bcs = [0.012 - (-0.47) μV/°C
    = 0.482 μV/°C2
    Putting this value in eq. (1), we get:
    Ecs = 46.46 × 10-6 × 40 + `` \frac{1}{2}`` × 0.482 × 10-6 × (40)2
    = 1.04 × 10-5 V
    Page No 219:
  • Qstn #15
    Find the neutral temperature and inversion temperature of a copper-iron thermocouple if the reference junction is kept at 0°C. Use the data in the table (33.1).
    Ans : Neutral temperature,
    `` {\theta }_{n}=-\frac{a}{b}``
    `` {a}_{CuFe}={a}_{CuPb}-{a}_{FePb}``
    `` =2.76-16.6=13.84\,\mathrm{\,\mu V\,}°{\,\mathrm{\,C\,}}^{-1}``
    `` {b}_{CuFe}={b}_{CuPb}-{b}_{FePb}``
    `` =0.012+0.030=0.042\,\mathrm{\,\mu V\,}°{\,\mathrm{\,C\,}}^{-2}``
    Thus, the neutral temperature,
    `` {\theta }_{n}=\frac{-{a}_{CuFe}}{{b}_{CuFe}}=\frac{13.84}{0.042}=329.52°\,\mathrm{\,C\,}=330°\,\mathrm{\,C\,}``
    The inversion temperature is double the neutral temperature, i.e. 659 `` °``C.
    Page No 219:
  • Qstn #16
    Find the charge required to flow through an electrolyte to liberate one atom of
  • #16-a
    a monovalent material and
    Ans : Amount of charge required by 1 equivalent mass of the substance = 96500 C
    For a monovalent material,
    equivalent mass = molecular mass
    ⇒ Amount of charge required by 6.023 × 1023 atoms = 96500 C
    ∴ Amount of charge required by 1 atom = `` \frac{96500}{6.023\times {10}^{23}}=1.6\times {10}^{-19}\,\mathrm{\,C\,}``
  • #16-b
    a divalent material.
    Ans : For a divalent material,
    equivalent mass =`` \frac{1}{2}``molecular mass
    ⇒ Amount of charge required by `` \frac{1}{2}`` × 6.023 × 1023 = 96500 C
    ∴ Amount of charge required by 1 atom = 1.6 × 2 × 10-19 = 3.2 × 10-19 C
    Page No 219:
  • Qstn #17
    Find the amount of silver liberated at the cathode if 0.500 A of current is passed through an AgNO3 electrolyte for 1 hour. Atomic weight of silver is 107.9 g mol-1.
    Ans : Equivalent mass of silver, EAg = 107.9 g (∵ Ag is monoatomic)
    The ECE of silver,
    `` {Z}_{\mathit{A}\mathit{g}}=\frac{{E}_{Ag}}{f}=\frac{107.9}{96500}=0.001118``
    Using the formula, m = Zit, we get:
    m = 0.00118 × 0.500 × 3600
    = 2.01 g
    So, 2.01 g of silver is liberated.
    Page No 219:
  • Qstn #18
    An electroplating unit plates 3.0 g of silver on a brass plate in 3.0 minutes. Find the current used by the unit. The electrochemical equivalent of silver is 1.12 × 10-6 kg C-1.
    Ans : Given:
    Mass of silver deposited, m = 3 g
    Time taken, t = 3 min. = 180 s
    E.C.E. of silver, Z = 1.12 × 10-6 kg C-1
    Using the formula, m = Zit, we get:
    `` 3\times {10}^{-3}=1.12\times {10}^{-6}\times i\times 180``
    `` ``
    `` \Rightarrow i=\frac{3\times {10}^{-3}}{1.12\times {10}^{-6}\times 180}``
    `` \Rightarrow i=\frac{1}{6.72}\times {10}^{2}``
    `` \Rightarrow i=14.89\approx 15\,\mathrm{\,A\,}``
    Page No 219:
  • Qstn #19
    Find the time required to liberate 1.0 litre of hydrogen at STP in an electrolytic cell by a current of 5.0 A.
    Ans : Let the required time be t.
    Mass of 1 litre hydrogen,
    `` m=\frac{2}{22.4}\,\mathrm{\,g\,}``
    `` ``
    Using the formula, m = Zit, we get:
    `` \frac{2}{22.4}=\frac{1\times 5\times t}{96500}``
    `` \Rightarrow t=\frac{2\times 96500}{22.4\times 5}``
    `` \Rightarrow t=1723.21\,\mathrm{\,s\,}=28.7\,\mathrm{\,minutes\,}\approx 29\,\mathrm{\,minutes\,}``
    Page No 219:
  • Qstn #20
    Two voltameters, one with a solution of silver salt and the other with a trivalent-metal salt, are connected in series and a current of 2 A is maintained for 1.50 hours. It is found that 1.00 g of the trivalent metal is deposited.
    Ans : Given:
    Mass of salt deposited, m = 1 g
    Current, i = 2 A
    Time, t = 1.5 hours = 5400 s
    For the trivalent metal salt:
    Equivalent mass = `` \frac{1}{3}``Atomic weight
    The E.C.E of the salt,
    `` Z=\frac{\,\mathrm{\,Equivalent\,}\,\mathrm{\,mass\,}}{96500}=\frac{\,\mathrm{\,Atomic\,}\,\mathrm{\,weight\,}}{3\times 96500}``
  • #20-a
    What is the atomic weight of the trivalent metal?
    Ans : Using the formula, m = Zit, we get:
    `` 1\times {10}^{-3}=\frac{\,\mathrm{\,Atomic\,}\,\mathrm{\,weight\,}}{3\times 96500}\times 2\times 5400``
    `` \Rightarrow \,\mathrm{\,Atomic\,}\,\mathrm{\,weight\,}=\frac{3\times 96500\times {10}^{-3}}{2\times 5400}=26.8\times {10}^{-3}\,\mathrm{\,kg\,}/\,\mathrm{\,mole\,}``
    `` \Rightarrow \,\mathrm{\,Atomic\,}\,\mathrm{\,weight\,}=26.8\,\mathrm{\,g\,}/\,\mathrm{\,mole\,}``
  • #20-b
    How much silver is deposited during this period? Atomic weight of silver is 107.9 g mol-1.
    Ans : Using the relation between equivalent mass and mass deposited on plates, we get:
    `` \frac{{E}_{\mathit{1}}}{{E}_{\mathit{2}}}=\frac{{m}_{1}}{{m}_{2}}``
    `` \Rightarrow \frac{26.8}{3\times 107.9}=\frac{1}{{m}_{2}}``
    `` \Rightarrow {m}_{2}=12.1\,\mathrm{\,g\,}``
    `` ``
    Page No 219:
  • Qstn #21
    A brass plate of surface area 200 cm2 on one side is electroplated with 0.10 mm thick silver layers on both sides using a 15 A current. Find the time taken to do the job. The specific gravity of silver is 10.5 and its atomic weight is 107.9 g mol-1.
    Ans : Given:
    Current, i = 15 A
    Surface area of the plate = 200 cm2,
    Thickness of silver deposited= 0.1 mm = 0.01 cm
    Volume of Ag deposited on one side = 200 × 0.01 cm3 = 2 cm3
    ∴ Volume of Ag deposited on both side = 4 cm3
    Mass of silver deposited,
    m = Volume × Specific gravity × 1000 = 4 × 10-3 × 10.5 ×1000 = 42 kg
    Using the formula, m = Zit, we get:
    42 = ZAg × 15 × t
    `` \Rightarrow t=\frac{42\times 96500}{107.9\times 15}\,\mathrm{\,s\,}``
    `` \Rightarrow t=2504.17\,\mathrm{\,s\,}=42\,\mathrm{\,minutes\,}``
    Page No 219:
  • Qstn #22
    The figure shows an electrolyte of AgCl through which a current is passed. It is observed that 2.68 g of silver is deposited in 10 minutes on the cathode. Find the heat developed in the 20 Ω resistor during this period. Atomic weight of silver is 107.9 g mol-1.
    Figure
    Ans : Given:
    Mass of silver deposited, m = 2.68 g
    Time, t = 10 minutes = 600 s
    Using the formula, m = Zit, we get:
    `` 2.68\times {10}^{-3}=\frac{107.9\times {10}^{-3}}{96500}\times i\times 600``
    `` \Rightarrow i=\frac{2.68\times 96500}{107.9\times 600}``
    `` \Rightarrow i=3.99=4\,\mathrm{\,A\,}``
    Heat developed in the 20 Ω resistor,
    `` H={i}^{2}Rt``
    `` \Rightarrow H={\left(4\right)}^{2}\times 20\times 600``
    `` \Rightarrow H=192000\,\mathrm{\,J\,}=192\,\mathrm{\,kJ\,}``
    Page No 219:
  • Qstn #23
    The potential difference across the terminals of a battery of emf 12 V and internal resistance 2 Ω drops to 10 V when it is connected to a silver voltameter. Find the silver deposited at the cathode in half an hour. Atomic weight of silver is 107.9 g mol-1.
    Ans : Let i be the current through the circuit.
    Emf of battery, E = 12 V
    Voltage drop across the voltameter, V = 10 V
    Internal resistance of the battery, r = 2 Ω
    Applying Kirchoff's Law in the circuit, we get:
    `` E=V+ir``
    `` \Rightarrow i=\frac{E-V}{r}=\frac{12-10}{2}=1\,\mathrm{\,A\,}``
    Using the formula m = Zit, we get:
    `` m=\frac{107.9}{96500}\times 1\times 0.5\times 3600=2.01\,\mathrm{\,g\,}``
    Page No 219:
  • Qstn #24
    A plate of area 10 cm2 is to be electroplated with copper (density 9000 kg m-3) to a thickness of 10 micrometres on both sides, using a cell of 12 V. Calculate the energy spent by the cell in the process of deposition. If this energy is used to heat 100 g of water, calculate the rise in the temperature of the water. ECE of copper = 3 × 10-7 kg C-1 and specific heat capacity of water = 4200 J kg-1
    Ans : Surface area of the plate, A = 10 cm2 = 10 × 10-4 m2
    Thickness of copper deposited, t = 10 μm = 10-5 m
    Density of copper = 9000 kg/m3
    Volume of copper deposited, V = A(2t)
    V = 10 × 10-4 × 2 × 10 × 10-6
    = 2 × 102 × 10-10
    = 2 × 10-8 m3
    Mass of copper deposited, m = Volume × Density = 2 × 10-8 × 9000
    ⇒ m = 18 × 10-5 kg
    Using the formula, m = ZQ, we get:
    18 × 10-5 = 3 × 10-7 × Q
    ⇒ Q = 6 × 102 C
    Energy spent by the cell = Work done by the cell
    ⇒W = VQ
    = 12 × 6 × 102
    = 72 × 102 = 7.2 kJ
    Let ∆θ be the rise in temperature of water. When this energy is used to heat 100 g of water, we have:
    7.2 × 103 = 100 × 10-3 × 4200 × ∆θ
    ⇒ ∆θ = 17 K
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